Electrochemistry

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Electrochemistry

  1. 1. M e t a l l u r g i c a l E n g i n e e r i n g D e p a r t m e n t 0ctober 1st, 2013 Chapter 4 ASSIGNMENTNUMERICALSLECTURES ELECTROCHEMISTRY CY-105 APPLIED CHEMISTRY
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  3. 3. Page 3 of 50 Muhammad Umair Tariq (MY-065) Syed Muhammad Tanveer (MY-017) Anas Hanif (MY-037) Ahmed Mazher (MY-011) Raheel Nadeem (MY-059) Ali Fakhar (MY-066) Danish Khan (MY-069) Course Instructor: SIR SAJJAD HAIDER. SECTIONS: 1. THEORY SECTION. 2. CLASS AND HOMEWORK NUMERICALS. 3. ASSIGNMENT NUMERICALS. 4. IMPORTANT KEY POINTS AND QUESTIONS. 5. M.C.Qโ€™s RELATING TO ELECROCHEMISTRY.
  4. 4. Page 4 of 50 Electrochemistry Electrolysis Mechanism of Electrolysis Electrochemical Cell Half Cell , VoltaicCell Corrosion and Inhibation of corrosion Faraday's Law Of Electrolysis
  5. 5. Page 5 of 50 1. THEORY SECTION Electrochemistry It is the study of interconversion between chemical and electrical energy. IMPORTANT TERMINOLOGIES USED IN ELECTROCHEMISTRY Before we start the study of electrochemistry, we must be familiar with a few common terms: i. Current is the flow of electrons through a wire or any conductor . ii. Electrode is the material (a metallic rod/bar which conducts electrons into and out of a solution. iii. Anode is the electrode at which oxidation occurs. It sends electrons into the outer circuit. It has negative charge and is shown a (-) in cell diagram. iv. Cathode is the electrode at which electrons are received from the outer circuit. It has a positive charge and is shown as (+) in cell diagrams. v. Anode cell is the cell of the in which oxidation half reaction occurs. It contains the anode vi. Cathode cell is the cell in which reduction half reaction occurs it contains the e cathode. vii. Oxidation is the loss of electrons OR gain of O2 OR loss of H+ OR increase of oxidation number. viii. Reduction is the gain of electrons OR loss of O2 ORgain of H+ OR increase of reduction number. ix. Reducing agent are those which donates the electrons and is oxidized. x. Oxidizing agent are those which accepts electrons and is reduced. xi. Electrolyte is a compound in a solution or a molten compound which conducts electricity and is decomposed by it in the process. xii. Non-Electrolyte is a solution of a compound or a molten compound which does not conduct electricity or become decomposed by it. xiii. Salt-bridge in electrochemistry is a laboratory device used to connect the oxidation and reduction half cells of a galvanic cell ( voltaic cell ) a type of electrochemical cell. xiv. Electromotive force is the energy per unit charge that is converted reversibly from chemical, mechanical or other forms of energy into electrical energy in a battery or dynamo.
  6. 6. Page 6 of 50 ELECTROLYSIS It is the chemical decomposition of a compound brought about by the passage of direct current through solution of the compound or a molten compound. MECHANISM OF ELECTROLYSIS Before providing the current to the electrolytic cell the ions moves randomly in the solution. When the current is applied to the cell. The an-ion of the electrolyte moves towards the anode and cat ion towards the cathode. Oxidation take place to the anode where electrons are lost and gain of electron (reduction) takes place at cathode. Note: Anions are negatively charged that is why it moves towards anode which is positive and cat-ions are positively charged hence move towards cathode. EXAMPLE: Consider electrolyte of HCl. HCl โ†’ ๐ป+ + ๐ถ๐‘™โˆ’ Here hydrogen ion moves towards cathode and chlorine ion moves toward anode. At cathode: ๐ป+ + ๐‘’โˆ’ โ†’ H Each hydrogen ion picks up an electron from the cathode to become a hydrogen gas. At anode: ๐ถ๐‘™โˆ’ โ†’ Cl + ๐‘’+ After the chloride ion loses its electron to the anode pair of chlorine atom unit to form chlorine gas.
  7. 7. Page 7 of 50 DIFFERENCES 1. Electrolyte and Non-Electrolyte 2. Insulator and Non-Electrolyte INSULATOR NON-ELECTROLYTE They do not conduct electricity. They do not conduct electricity. They are solid They are not solid Example: Plastic Example: Covalent liquids or organic liquids. 3. Conductor and Electrolytes CONDUTOR ELETROLYTES They conduct electricity due to electrons. They conduct electricity due to ions. They are solid substances. They are not solid substance either liquid (solution) or molten. It does not involve any chemical reaction. It involves chemical reaction. They are not change in any way by supplying electricity They may change in any way by supplying electricity. EXAMPLE : metal, graphite. EXAMPLE: acid, alkalis, salt solution. ELECTROLYTE NON-ELECTROLYTE It is the ionic substances. It is the covalent substance. It conducts electricity in a solution or in a molten state. It does not conduct electricity in a solution or molten state. They are inorganic compounds. They are mainly organic compounds. For example: Urea,benzene, trichloro methane etc. For example: Solution of sodium chloride, copper sulphates etc.
  8. 8. Page 8 of 50 STRONG ELECTROLYTE: A strong electrolyte is an electrolyte that completely dissociates in solution. The solution will contain only ions and no molecules of the electrolyte. Strong electrolytes are good conductors of electricity. Examples: HCl (hydrochloric acid), H2SO4 (sulfuric acid), NaOH (sodium hydroxide) and KOH (potassium hydroxide) are all strong electrolytes. Strong electrolytes conduct current very efficiently. Completely ionized or dissociate when dissolved in water. a. Soluble ionic compounds. b. Strong acids (HNO3(aq) ,H2SO4 (aq) , HCl(aq)) HNO3 โ†’ H+ + NO3 - (100% ionization) c. Strong bases (KOH and NaOH) KOH โ†’ K+ + OH- (100% dissociation) WEAK ELECTROLYTE : A weak electrolyte is an electrolyte that does not completely dissociate in solution. The solution will contain both ions and molecules of the electrolyte. Examples: HC2H3O2 (acetic acid), H2CO3 (carbonic acid), NH3 (ammonia) are all weak electrolytes a. Weak acids (organic acids โ†’ acetic, citric, butyric, malic etc.) HC2H3O2 โ†’ H+ + C2H3O2 - b. Weak bases (ammonia) NH3 + H2O โ†’ NH4 + OH-
  9. 9. Page 9 of 50 Laws of Electrolysis Michael Faraday Michael Faraday, was an English scientist who contributed to the fields of electromagnetism and electrochemistry. His main discoveries include those of electromagnetic induction, diamagnetism and electrolysis. INTRODUCTION: Faraday's laws of electrolysis are quantitative relationships based on the electrochemical researches published by Michael Faraday in 1833. Faradayโ€™s Laws of Electrolysis give the relationship between the amount of material liberated at the electrode and the amount of electric energy that is passed through the electrolyte. Faraday's work on the chemical reaction produced when an electric current passes through a liquid resulted in the laws of electrolysis. There are 2 laws of Electrolysis: 1- Faraday's first law of electrolysis. 2- Faraday's second law of electrolysis. Faraday's laws of electrolysis are quantitative relationships based on the electrochemical researches published in 1833. Faraday's first law of electrolysis: Statement: โ€œThe amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of current passing through the electrolyteโ€ Explanation: It means that the mass of a substance produced by electrolysis is proportional to the quantity of electricity used. In order to produce more substance electrolytically we must use more electricity. m ฮฑ Q m ฮฑ It m=zIt where โ€œZโ€ is known as electrochemical equivalent.
  10. 10. Page 10 of 50 Electrochemical Equivalent โ€œzโ€: โ€œIf the mass is 1 kg and current of 1 ampere is passed through electrolyte in 1 second then the mass deposited or liberated is called 1 electrochemical equivalent โ€ OR โ€œThe mass deposited by 1 coulomb charge is called 1 electrochemical equivalentโ€ z=m / It or z=m/Q For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited is 0.001118 g. this is the value of electrochemical equivalent of silver. Importance of Faradayโ€™s First law of Electrolysis: With the help of faradayโ€™s First law we are able to calculate: 1) The value of electrochemical equivalents of different substances. 2) The masses of substances produced by passing a known quantity of electricity through their solution. Faradayโ€™s Second Law of Electrolysis: โ€œIt states that masses of different substances deposited or liberated when same quantity of current is passed through different electrolytes connected in series are proportional to their chemical equivalent massesโ€ Explanation: Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of electricity through them for the same time. Now Ag, Cu and Al metals collect at the cathode. Their masses are directly proportional to their equivalent masses. According to Faraday, if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes, we get which are the equivalent masses of Ag, Cu and Al respectively. Ag+ 1e- Ag(S) Cu2+ + 2e- Cu (S) Al+3 +3e- Al(S) Charge on 1 electron = 96500 coulomb = 1 faraday
  11. 11. Page 11 of 50 Equivalent Mass: โ€œThe mass deposited or liberated which is equal to 108 parts of silver, 35.5 parts of chlorine and 8 parts of oxygenโ€ . Importance of Faradayโ€™s Second Law: By the help of 2nd law we can calculate: ๏‚ท Equivalent weight of metal. ๏‚ท Unit of electric charge (coulomb). ๏‚ท Avogadroโ€™s number. Steps for doing numerical: Step#1: Find Charge OR Coulomb Step #2: Find Faraday Step #3: Taking the calculated values of Charge and faraday, solve the Question further for given equations or for given solutions.
  12. 12. Page 12 of 50 2. Class Numericals Q) A direct current of 0.01 ampere flows for 4 hours through 3 cells in series. They contain AgNO3, CuSO4 and AuNO3. Calculate the mass deposited in each cell. SOLUTION: Step#1: Find Charge OR Coulomb q=It = (0.01)(4 x 60 x 60) q= 144 coulomb Step #2: Find Faraday 96500 coulomb = 1 Faraday 144 coulomb = 1/96500 x 144 F= 1.49 x 10-3 Farad. Step #3: Taking the calculated values of Charge and faraday, solve the question further for given equations Let X be the required amount of AgNO3, CuSO4 and AuNO3. Ag+ 1e- โ†’ Ag(S) 108 g = 1 F 1.49 x 10-3 = X gm X= 0.161 gm. Cu2+ + 2e- โ†’ Cu(S) 63.5/2= 1 F OR 2 F= 63.5 gm 1.49 x 10-3 = X 2X = 63.5 x (1.49 x 10-3)/2 X= 0.0473 gm. Al+3 +3e- ๐ด๐‘™(๐‘ ) 197 gm = 3 F 3 F= 197 gm 1.49 x 10-3 F= X 3X= 197 x (1.49 x 10-3)/3 X= 0.0983 gm
  13. 13. Page 13 of 50 Q) A metal Ar=27 is deposited by electrolysis when 0.15 amp flows 3.5 hours, 0.176gm of metal is deposited. What is the charge in the cation of this metal? Solution: Step#1: Find Charge OR Coulomb q = It =0.15 (3.5x60x60) q= 1890 coulomb Step #2: Find Faraday 96500 coulomb = 1 faraday 1890 coulomb = x x= 1890/96500 x= 0.0196F Step#3: 0.176gm= 0.0196F 27gm = x faraday x = 0.0196ร—27/0.176 x = 3 charge is +3 Q) How many grams of oxygen gas liberated when 0.0565 ampere current passed during 185 seconds during electrolysis of water. Solution: Step#1: Find Charge OR Coulomb q=It = (0.0565)(185) =10.452 coulomb
  14. 14. Page 14 of 50 Step #2: Find Faraday 1F= 96500 Coulomb 10.452 C= 10.452/ 96500 = 1.083 x 10-4 F Step#3: 1 F = 8 gm 1.083 x 10-4 F= X gm X= 0.000864 gm Q) Find the amount of current if 404mg Cu is deposited during 5 hours. Electrolyte is CuSO4. q= It I= q/t -------------(1) Cu2+ + 2e- Cu(S) 63.5 gm = 2F 404 x 10-3 = X farad X 63.5 gm = 8.08 x 10 -3 X = 1.272 x 10 -4 F 1 F= 96500 Coulomb 1.272 x 10 -4 F= 96500 x 1.272 x 10 -4 F 12.272 coulomb I= q/t = 12.272 / 5x60x60= 0.0680 Ampere. HALF REACTION Let us consider the reaction 2๐‘๐‘Ž + ๐ถ๐‘™2 โ†’ 2๐‘๐‘Ž+ + ๐ถ๐‘™โˆ’ It occur by the transfer of electron from Na to Cl. Na lose an electron and it said to be oxidized to Na+ ion at the same time Cl gain a electron and is reduced to ๐ถ๐‘™โˆ’ ion. Such a reaction in which loss of electron (oxidation) and gain of elctron (reduction) occurs simultaneously, is called an Oxidation Reduction reaction or Redox Reaction. The redox reaction can be considered as made up of two reactions. For examples the redox reaction 2๐‘๐‘Ž + ๐ถ๐‘™2 โ†’ 2๐‘๐‘Ž+ + ๐ถ๐‘™โˆ’ is composed of two reactions 2๐‘๐‘Ž โ†’ 2๐‘๐‘Ž+ + 2๐‘’โˆ’ (Oxidation) ๐ถ๐‘™2 + 2๐‘’โˆ’ โ†’ 2๐ถ๐‘™โˆ’ (Reduction)
  15. 15. Page 15 of 50 Each of the two reactions shows just oxidation or just the reduction portion of the overall redox reaction. These two components of redox reactions are called Half reactions. The first half reaction that proceeds by oxidation is often referred to as the Oxidation Half reaction. The second half reaction that occurs by reduction is referred to as the Reduction Half reaction. When the two half reactions are added together the sum is the net redox reaction. ELECTROCHEMICAL CELLS A device used for producing an electrical current from a chemical reaction (redox reaction) is called an electrochemical cells. Explanation: When we read the above defination, the first question which came into our minds is how a redox reaction can produce an electrical current? So we perform an experiment and check it. Let consider a beaker containing ๐ถ๐‘ข๐‘†๐‘‚4 and a zinc rod. When bar of zinc is dipped in the solution of copper sulphates redox reaction taking place on zinc bar and copper metal is deposited on the bar. Now there is a burning question i.e. why Cu is reduced on zinc bar and why we use Zn bar? We may use Cu bar dipped in ๐‘๐‘›๐‘†๐‘‚4 solution? The answer is very simple. Cu is reduced because Cu has ability to reduced but this answer is not complete, the real reason is Cu has greater redox potential value than Zn. Whenever reduction occurs deposition takes place only and only at cathode. If we dipped Cu bar in ๐‘๐‘›๐‘†๐‘‚4 no reaction occurs. Zn do not depositon Cu bar because it has less pedox potential value than Cu. The two half reactions are ๐ถ๐‘ข+2 + 2๐‘’โˆ’ โ†’ ๐ถ๐‘ข (reduction) ๐‘๐‘› โ†’ ๐‘๐‘›+2 + 2๐‘’โˆ’ (oxidation) The net reaction is
  16. 16. Page 16 of 50 ๐ถ๐‘ข2+ + ๐‘๐‘› โ†’ ๐ถ๐‘ข( ๐‘ ) + ๐‘๐‘›( ๐‘Ž๐‘ž) 2+ In this change Zn is oxidized to give ๐‘๐‘›2+ ions and ๐ถ๐‘ข2+ are reduced to Cu atoms. The electrons released in the first half reaction are used up by the second half reaction. Both the half reaction occurs on the zinc rod itself and there is no net charge. Now let the two half reaction occurs in separate compartments which are connected by a wire. The electrons produced in the left compartment flow through the wire to the other compartment. However, the current will flow for an instant then stop. The current stops flowing because of the charge built up in the two cells. The electrons leave the left compartments and it would become positively charged. The right cell receives electrons and become negatively charged both these factors oppose the flow of electrons which eventually stops. This problem is solved by connecting a salt bridge (u shaped tube containing electrolyte) between the two cells. It provides a passage of ion from one cell to the other cell. With this flow of ion, circuit is complete and electrons pass freely through the wire to keep the net charge zero in the two compartments. VOLTAIC CELL
  17. 17. Page 17 of 50 A voltaic cell also known as a Galvonic Cell is a electrochemical cell in which electrical current is generated by a spontaneous redox reaction. Explanation: Consider a electrochemical cell in which a rod of zinc (anode) is placed in ๐‘๐‘›๐‘†๐‘‚4 solution in the cell A. A rod of Cu (cathode) is immersed in ๐ถ๐‘ข๐‘†๐‘‚4 solution in the cell B. The Zn and Cu electrodes are joined by a copper wire .When the cell is set up, electrons flow from Zn electrode through the wire to the Cu cathode. As a result, IN CELL A At anode half cell, redox reaction takes place. Zn dissolves in the solution to form more and more and more ๐‘๐‘›+2 ions then at a certain time movement of ๐‘’โˆ’ stops since current stops. IN CELL B At cathode half cell more and more ๐‘’โˆ’ comes from cell A . The Cu ions in the cathode half cell pick up electrons and are converted to Cu atoms on the cathode. Then at a certain time income of ๐‘’โˆ’ stops since current in cell B also stops. In cell A ๐‘๐‘›+2 stops the moment of ๐‘’โˆ’ and in cell B ๐‘†๐‘‚4 โˆ’ stops income of ๐‘’โˆ’ .So we construct a Salt Bridge. ๐‘๐‘›2+ + ๐‘†04 โˆ’ โ†’ ๐‘๐‘›๐‘†๐‘‚4 Cell B Cell A
  18. 18. Page 18 of 50 ELECTROLYSIS OF BRINE Brine is NaCl (concentrated) i.e. low H2O and high Na. Brine is a concentrated or saturated solution of sodium chloride in water. The process of electrolysis uses electrical energy to produce a chemical reaction in the sodium chloride solution which results in the formation of sodium hydroxide, chlorine and hydrogen gas. The equation for this chemical reaction is as follows. Sodium chloride + Water โ†’ Sodium hydroxide + Hydrogen + Chlorine NaCl + H2O โ†’ NaOH + H2 + Cl2 The manufacture of sodium hydroxide or caustic soda is an important industrial process. The electrochemistry involved in the reaction involves an reduction oxidation reaction, which is sometimes called a redox reaction. In a solution of brine there are three chemical species: The sodium ion, Na+, the chloride ion, Cl- and water, H2O. The easiest chemcial species to oxidise is the chloride ion. The easiest chemical species to reduce is the water. Reaction at anode: The anode is the electrode where oxidation or the losses of electrons take place. In an electrolytic cell the anode is positive in charge. The chloride ion is oxidized to chlorine gas. 2Cl- (aq) โ†’ Cl + 2e- OH- โ†’ ยฝH2 + O2(g). Reaction at cathode: Na+ (aq) + e- โ†’ Na(s) In water the sodium ion is not reduced as it is much easier to reduce water. The sodium ion becomes a spectator ion and is not involved in the reaction. Overall reaction: The overall chemical equation for the electrolysis of brine is NaCl + H2O โ†’ H2 + Cl2 + NaOH GOOD TO KNOW: In industry the hydrogen gas must be separated from the chlorine gas or they will explode when exposed to UV light producing hydrogen chloride. ๐ป2 + ๐ถ๐‘™2 โ†’ 2๐ป๐ถ๐‘™ Sodium hydroxide is caustic in nature and must be handled with care. It is used in industry in the manufacture of soap by saponification. It is also used to make pulp and in the production of paper.
  19. 19. Page 19 of 50
  20. 20. Page 20 of 50 CORROSION Corrosion is an electrochemical reaction composed of two half cell reactions, an anodic reaction and cathodic reaction. In anodic reactions electron releases and in cathodic reactions electron consumes. An electrochemical reaction is defined as a chemical reaction involving the transfer of electrons. It is also a chemical reaction which involves oxidation and reduction. Since metallic corrosion is almost always an electrochemical process, it is important to understand the basic nature of electrochemical reactions. The discoveries that gradually evolved in modern corrosion science have, in fact, played an important role in the development of a multitude of technologies we are enjoying today. An important achievement early in the history of electrochemistry was the production of power sources, following the production of the first batteries by Alessandro Volta illustrates the principle of a Daniel cell in which copper and zinc metals are immersed in solutions of their respective sulfates. The Daniel cell was the first truly practical and reliable electric battery that supported many nineteenth-century electrical innovations such as the telegraph. In the process of the reaction, electrons can be transferred from the corroding zinc to the copper through an electrically conducting path as a useful electric current. Zinc more readily loses electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow through an external wire which leads from the zinc to the copper. The difference in the susceptibility of two metals to corrode can often cause a situation that is called galvanic corrosion named after Luigi Galvani, the discoverer of the effect. The purpose of the separator shown in is to keep each metal in contact with its own soluble sulfates, a technical point that is critical in order to keep the voltage of a Daniel cell relatively constant . The same goal can be achieved by using a salt bridge between two different beakers. The salt bridge in that case, provides the electrolytic path that is necessary to complete an electrochemical cell circuit. This situation is common in natural corrosion cells where the environment serves as the electrolyte that completes the corrosion cell. The conductivity of an aqueous environment such as soils, concrete, or natural waters has often been related to its corrosivity. Zinc anode: Zn(s)โ†’ Zn+2+2e- Copper cathode: Cu2+ โ†’ 2e- + Cu(s)
  21. 21. Page 21 of 50 The short-hand description is valid for both cells. Such a description is often used to simplify textual reference to such cells. Zn/Zn2๏€ซ, SO24 (Conc1)//Cu2๏€ซ, SO24 (Conc2))/Cu(๏€ซ) Conc1 andConc2 indicate respectively the concentration of zinc sulfate and copper sulfate that may differ in the two half-cells while the two slanted bars (//) describe the presence of a separator. The same short-hand description also identifies the zinc electrode as the anode that is negative in the case of a spontaneous reaction and the copper cathode as positive. The fact that corrosion consists of at least one oxidation and one reduction reaction is not always as obvious as it is in chemical power cells and batteries. The two reactions are often combined on a single piece of metal as it is illustrated schematically A piece of zinc immersed in hydrochloric acid solution is undergoing corrosion. At some point on the surface, zinc is transformed to zinc ions. This reaction produces electrons and these pass through the solid conducting metal to other sites on the metal surface where hydrogen ions are reduced to hydrogen gas Anodic reaction: Zn(s)โ†’2e- + Zn+2 Cathodic reaction: 2H+2+2e- โ†’ H2(g) The nature of an electrochemical reaction typically illustrated for zinc .During such a reaction, electrons are transferred, or, viewing it another way, an oxidation process occurs together with a reduction process. The overall corrosion processes are summarized in Overall corrosion reaction: Zn ๏€ซ2H๏€ซ ๏€  Zn2๏€ซ๏€ ๏€ซ๏€ H2(g) Briefly then, for corrosion to occur there must be a formation of ions and release of electrons at an anodic surface where oxidation or deterioration of the metal occurs. There must be a simultaneous reaction at the cathodic surface to consume the electrons generated at the anode. These electrons can serve to neutralize must go on at the same time and at equivalent rates. However, what is usually recognized as the corrosion process occurs only at the areas that serve as anodes. Anodic Processes: Let us consider in greater detail what takes place at the anode when corrosion occurs. For instance, reconsider. This re- action involves the reduction of hydrogen ions to hydrogen gas,. This hydrogen evolution reaction occurs with a wide variety of metals and acids, including hydrochloric, sulfuric, perchloric, hydrofluoric, formic, and other strong acids. The individual anodic reactions for iron, nickel, and aluminum are listed as follows: Iron anodic reaction: Fe(s) Fe2๏€ซ ๏€  ๏€ซ๏€ 2e Nickel anodic reaction: Ni(s) Ni2๏€ซ ๏€  ๏€ซ๏€ 2e Aluminum anodic reaction: Al(s) Al3๏€ซ ๏€  ๏€ซ๏€ 3e That is, the corrosion of metal M results in the oxidation of metal M to an ion with a valence charge of nand the release of n electrons. Some metals such as silver are univalent, while other metals such as iron, titanium, and uranium are multivalent and possess positive charges as high as is general and applies to all corrosion reactions.
  22. 22. Page 22 of 50 Cathodic Process When hydrogen ions are reduced to their atomic form they often combine, as shown earlier, to produce hydrogen gas through reaction with electrons at a cathodic surface. This reduction of hydrogen ions at (OH ) ions and make the solution less acidic or more alkaline or basic at the corroding interface. In neutral waters the anodic corrosion of some metals like aluminum, zinc, or magnesium develops enough energy to split water directly. Water splitting cathodic reaction: 2H2O(l) ๏€ซ๏€ 2e- H2 ๏€ซ๏€ 2OH- The change in the concentration of hydrogen ions or increase in hydroxyl ions can be shown by the use of pH indicators, which change color and thus can serve to demonstrate and locate the existence of surfaces on which the cathodic reactions in corrosion are taking place. There are several other cathodic reactions encountered during the corrosion of metals. These are listed below: Oxygen reduction: (acid solutions) O2 ๏€ซ๏€ 4H๏€ซ๏€ ๏€ซ๏€ 4e 2H2O (neutral or basic solutions) O2 ๏€ซ๏€ 2H2O ๏€ซ๏€ 4e 4OH Hydrogen evolution: 2H๏€ซ๏€ ๏€ซ๏€ 2e H2(g) Metal ion reduction: Fe3๏€ซ๏€ ๏€ซ๏€ e Fe2๏€ซ Metal deposition: Cu2๏€ซ๏€ ๏€ซ๏€ 2e Cu(s) Hydrogen ion reduction, or hydrogen evolution, has already been discussed. This is the cathodic reaction that occurs during corrosion in acids. Inhibition of corrosion Introduction: Inhibition means to remove corrosion. There are a number of ways to remove corrosion. Some methods are: ๏‚ท Impermeable layer. ๏‚ท Galvanizing ๏‚ท Cathodic production. ๏‚ท Sacrification method. Impermeable layer: Coating of impermeable layer on the material prevents it from corrosion. Sometimes this measure fails because of porous paint. Galvanizing: Coating of zinc on a material can also prevent it from being corroded. This method is called Galvanizing Cathodic production: This method was Invented in 1824 by the UK scientist Sir Humphrey Davy (1778-1829) to protect the copper-cladded wooden ships from seawater corrosion. It is an electrical method of preventing corrosion on metallic structures which are in electrolytes such as soil or water Sacrification method: This method is basically used in ships.Here Mg is used as a sacrificing agent. We attach Mg to Fe. First Mg oxidizes, and then Fe will oxidize after a long time.
  23. 23. Page 23 of 50 3. ASSIGNMENT NUMERICALS QUESTION # 1 Electrolysis involves the chemical decomposition of a compound either when molten or in aqueous solution, by the passage of an electric current. (a)Explain why aqueous calcium nitrate can be electrolyzed but liquid pentane cannot. ANSWER: Aqueous calcium nitrate can be electrolyzed because in aqueous form it contain ions which are responsible for electrolysis, but in liquid pentane ions are not formed that is why it cannot be electrolyzed. (B) State the products of electrolysis of molten sodium chloride. ANSWER: Electrolysis of molten sodium chloride yields sodium metal and chlorine gas.Sodium ions are reduced at cathode to sodium metal while chlorine gas is liberated at anode. (C) State the products of electrolysis of concentrated aqueous sodium chloride. ANSWER: Electrolysis of concentrated aqueous sodium chloride yields hydrogen and chlorine ( with aqueous sodium hydroxide remaining in solution. The reason for the difference is that the reduction of Na (E=2.7V) is energetically more difficult than reduction of water (-1.27V). (D) A student investigates the electrolysis of aqueous copper (II) sulphate using the apparatus shown below. . Copper anode (Positive electrode) Copper cathode (Negative electrode) Aqueous copper
  24. 24. Page 24 of 50 The student weighs the copper cathode before and after the electrolysis. Experiment Current used Time taken/s Mass of metal Before starting/g After electrolysis/g 1 2.0 180 1.24 1.36 2 4.0 180 1.20 1.44 3 2.0 360 1.34 1.58 (i)Explain with the help of an equation, why the cathode increases in mass. ANSWER: Cathode increases in mass in this reaction because reduction occurs at cathode and copper is deposited at cathode. Cathode reaction: ๐ถ๐‘ข2+ ( ๐‘Ž๐‘ž)+ 2๐‘’โˆ’ โ†’ ๐ถ๐‘ข (ii)In experiment 2 the student measures the mass of the anode both before and after the electrolysis. At the start the anode has a mass of 1.45g.Determine the mass of the anode at the end of the electrolysis. ANSWER: A we can see in experiment 2 the mass of cathode at the start is 1.20g and at the end it has a mass of 1.44g. Determine the mass of the anode at the end of the electrolysis, so taking the difference we get, 1.44-1.20=0.24g. Adding the given mass we get the required mass at the anode, 0.24+1.45=1.69g. (iii)The student does a fourth experiment, this time using a current of 8.0 A for 90 seconds. At the start the anode has a mass of 1.51g. Predict the mass of cathode at the end of the electrolysis. ANSWER: First we calculate the charge by using current and time, q=it q=8ร—90 q=720 C. For faraday, 96500C=1F 720C= 1 96500 ร— 720 =7.46ร— 10โˆ’3 F.
  25. 25. Page 25 of 50 Since, ๐ถ๐‘ข2+ + 2๐‘’ โ†’ ๐ถ๐‘ข 2F = 63.5g 7.46ร— 10โˆ’3 ๐น = x. X =0.236g. The total mass deposited on the cathode is calculated by adding the given mass in the calculated mass, X=0.236+1.51 . QUESTION # 2 Chlorine gas is manufactured by the electrolysis of brine using a diaphragm cell. (a)write half equations ,including state symbols, for the reactions occurring at each of the electrodes of a diaphragm cell. ANSWER: Reaction at cathode: ๐‘๐‘Ž+( ๐‘Ž๐‘ž) + 1e โ†’ Na ๐ป+ + 2๐‘’ โ†’ ๐ป2. Reaction at anode: 2๐ถ๐‘™โˆ’( ๐‘Ž๐‘ž) โ†’ ๐ถ๐‘™2(๐‘”) + 2๐‘’. ๐‘‚๐ปโˆ’ โ†’ ๐‘‚2 (b)In the diaphragm cell the anode is made of titanium and the cathode is made of steel. Suggest why steel is never used for anode. ANSWER: In electrolysis, the anodes are called as sacrificial anodes and they are usually made of zinc, aluminium and magnesium which are more electronegative than steel so are able to supply electrons to the electropositive steel because in electrolysis corrosion takes place so it is necessary to protect cathode so for cathodic protection anodes are not made of steel. (c)One important product made in the diaphragm cell is formed in the aqueous solution. (i)What substance is produced in aqueous solution in the diaphragm cell? X=1.746 gm
  26. 26. Page 26 of 50 ANSWER: Chlorine in water dissolves to produce HCL and HOCL being an oxidizing agent ,bleaches red litmus (d) Explain with the help of equations how this compound is formed by electrolysis? ANSWER: ๐ถ๐‘™2 + ๐ป2 ๐‘‚ โ†’ ๐ป๐ถ๐ฟ + ๐ป๐‘‚๐ถ๐‘™. HOCl + dye โ†’ ๐‘‘๐‘ฆ๐‘’ + ๐‘‚( ๐‘๐‘œ๐‘™๐‘œ๐‘ข๐‘Ÿ๐‘™๐‘’๐‘ ๐‘ ) + ๐ป๐ถ๐‘™. Question # 4 The electrolytic purification of copper can be carried out in an apparatus similar to the one shown below. + - The impure copper anode contains small quantities of metallic nickel, zinc and silver, together with inert oxides and carbon resulting from the initial reduction of the copper ore with coke. The copper goes into solution at the anode, but the silver remains as the metal and falls to the bottom as part of the anodeโ€™sludgeโ€™. The zinc also dissolves. (A)(1)Write a half equation including state symbols for the reaction of copper at the anode and precipitate pure copper onto the cathode. However, a small quantity of it is โ€˜wastedโ€™ in dissolving the impurities at the anode which then remain in solution. When a current of 20amp was passed through the cell for 10 hrs , it was found that 225 gm of pure copper was deposited on the cathode. At the cathode, copper ions are deposited as copper. At the anode, copper goes into solution as copper ions. (1)Part 1: Calculate the following. ๏‚ท Number of moles of copper produced at the cathode. Pure copper cathode CuSO4 (aqu.) Impure copper anode Anode โ€˜sludgeโ€™
  27. 27. Page 27 of 50 Answer : No. of moles are represented by = mass of element / molecular mass of element. So, no. of moles=225 / 63.5 = 3.54 mol. No. of moles of copper produced at cathode is 3.54 mol. ๏‚ท Number of moles of electrons that passed through the cell. 20 amp ร— 36000 secs = 720000 coulomb. 720000 C ร— 1F/96500 C = 7.4611 F. 7.46 F ร— 1mole e- /1F = 7.46 mole e- . So, 7.46 moles of electrons passed through the cell. ๏‚ท Number of moles of electrons needed to produce this copper. 7.46 mole of e- ร— 1 mol. Of copper/2 mole e- = 3.73 moles. So , 3.73 moles are needed to produce this copper. (1)Part 2: hence calculate the percentage of the current through the cell that has been โ€˜wastedโ€™ in dissolving the impurities at the anode. (2) Explain why silver remains as the metal. According to the electrochemical series copper (Cu) will deposit first and silver will remain as metal because silver at anode which is below copper in the electrochemical series (reactivity series) doesn't go into solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with any unreactive material left over from the ore. (3) Predict what happens to the nickel at the anode. Nickel according to the reactivity series will discharge into solution. (4) Write a half equation including state symbols for the main reaction at the cathode. AT CATHODE: Cu+2 + 2e- Cu. (5) Explain why zinc is not deposited on the cathode. Metals above copper in the electrochemical series (like zinc) will form ions at the anode and go into solution. However, they won't get discharged at the cathode provided their concentration doesn't get too high. The concentration of ions like zinc will increase with time, and the concentration of the copper ions in the solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion formed. (6) Suggest why the blue colour of the electrolyte slowly fades as the electrolysis proceeds. The blue colour of the electrolyte in cell fades when more and more Cu2+ ions are reduced to copper metal and plated onto the cathode as a pink deposit.Since copper ions in solution are used up, the blue colour fades Cu2+ + 2e- โ†’ Cu
  28. 28. Page 28 of 50 (B) Most of the current passed through the cell is used to dissolve the copper at the anode and precipitate pure copper onto the cathode. However,a small quantity of it is โ€˜wastedโ€™ in dissolving the impurities at the anode which then remain in solution. When a current of 20amp was passed through the cell for 10 hrs , it was found that 225 gm of pure copper was deposited on the cathode. (B)Part 1: Calculate the following. ๏‚ท Number of moles of copper produced at the cathode. Answer : No. of moles are represented by = mass of element / molecular mass of element. So, no. of moles=225 / 63.5 = 3.54 mol. No. of moles of copper produced at cathode is 3.54 mol. ๏‚ท Number of moles of electrons needed to produce this copper. 7.46 ole of e- ร— 1 mol. Of copper/2 mole e- = 3.73 moles. So , 3.73 moles are needed to produce this copper. ๏‚ท Number of moles of electrons that passed through the cell. 20 amp ร— 36000 secs = 720000 coulomb. 720000 C ร— 1F/96500 C = 7.4611 F. 7.46 F ร— 1mole e- /1F = 7.46 mole e- . So, 7.46 moles of electrons passed through the cell. (B)Part 2: Hence calculate the percentage of the current through the cell that has been โ€˜wastedโ€™ in dissolving the impurities at the anode. Question # 06 Define the following. (1) Electrochemical equivalent (Z) If the mass is 1kg and current of 1 amp pass through electrolyte in 1 second then the mass deposited or liberated is called 1 electro chemical equivalent. โ€˜ORโ€™ The mass deposited by 1 coulomb charge is called 1 electrochemical equivalent. Z = m/It ; Z= m / Q. (2) Standard electrode potential In an electrochemical cell, an electrode potential is created between two dissimilar metals. This potential is a measure of the energy per unit charge which is available from the oxidation/reduction reactions to drive the reaction. Also, the standard electrode potential is the potential difference between the electrode and an electrolyte (at 1 M), at standard conditions (1atm, 298K). It is represented by Eo. (3) Half reaction.
  29. 29. Page 29 of 50 A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Redox Reduction reaction Oxidation reaction. ยฝ oxidation reaction + ยฝ reduction reaction โ†’ redox reaction. For Example Ni(s) + 2๐น๐‘’3+ โ†’ ๐‘๐‘–2+ + 2๐น๐‘’2+ Oxidation takes place at the anode and the electrode must be Niโ†‘ ๐‘๐‘–2+ Ni(s)โ†’ ๐‘๐‘–2+ (๐‘Ž๐‘ž) + 2๐น๐‘’ And the reduction occurs at the cathode 2๐น๐‘’2+ +2eโ†’ 2๐น๐‘’2+ (4) Salt bridge. A salt bridge in electrochemistry is a laboratory device used to connect the oxidation and reduction half cells of a galvanic cell ( voltaic cell ) a type of electrochemical cell. (5) Electromotive force. The energy per unit charge that is converted reversibly from chemical ,mechanical or other forms of energy into electrical energy in a battery or dynamo. Question# 07 Define anti clockwise rule in electrolysis and if a direct current of 0.01amp flows for 4 hours through three cells in a series. They contain aqueous AgNO3 , CuSO4 and Au(NO3)3. Calculate the mass of metal deposited in each. (atomic masses. Ag=108, Cu=63.5, Au=197). ANTI CLOCKWISE RULE: It is a rule where positivity or negativity of an electrode is determined by considering the following steps. (1) Arrange w.r.t to redox potential values i.e the reactions should first be written in descending order of redox potential (Eo ) values. Cu+2 + 2e- โ†’ Cu. EO = 0.34 volts. Ag+ + e- โ†’ Ag. EO = 0.80 volts. (2) Make anti clockwise heads. Cu+2 + 2e- โ†’ Cu. EO = 0.34 volts. Ag+ + e- โ†’ Ag. EO = 0.80 volts. (3) The tail of arrow is the reactant and the head is product.
  30. 30. Page 30 of 50 For the mass of metal : Step#1:FindCharge OR Coulomb q=It = (0.01)(4 x 60 x 60) q= 144 coulomb Step#2: FindFaraday 96500 coulomb= 1 Faraday 144 coulomb= 1/96500 x 144 F= 1.49 x 10-3 Farad. Step#3: Takingthe calculatedvaluesof Charge andfaraday,solve the Questionfurtherforgivenequations Let X be the requiredamountof AgNO3,CuSO4 andAuNO3. Ag+ 1e- โ†’ Ag(S) 108 g = 1 F 1.49 x 10-3 = X gm X= 0.161 gm. Cu2+ + 2e- โ†’ Cu (S) 63.5/2= 1 F OR 2 F= 63.5 gm 1.49 x 10-3 = X 2X = 63.5 x (1.49 x 10-3 )/2 X= 0.0473 gm. Al+3 +3e- โ†’ Al(S) 197 gm = 3 F 3 F= 197 gm 1.49 x 10-3 F= X 3X= 197 x (1.49 x 10-3 )/3 X= 0.0983 gm
  31. 31. Page 31 of 50 Question # 08 Contruct the cell in which reactions are taken place, which of the electrodes shall act as anode (-ve electrode) and which one is cathode (+ve electrode). (a) Zn + CuSO4 โ†’ ZnSO4 + Cu (b) Cu + 2AgNO3 โ†’ Cu (NO3)2 + 2Ag (c) Zn + H2SO4 โ†’ ZnSO4 + H2 (d) Fe + SnCl2 โ†’ FeCl2 + Sn First we have to know that oxidation (loss of electrons) occurs at anode and reduction (gain of electrons) occurs at cathode. Oxidation: Zn ๏‚ฎ Zn2+ + 2 e- . Reduction: Cu2+ + 2 e ๏‚ฎ Cu Steps : 1) First write in ionic form. 2) Then write half cell reactions and balance them. 3) Then determine whether the electrode is cathode or anode. (a) Zn + CuSO4 โ†’ ZnSO4 + Cu. 1) Zn(s) + Cu+2 โ†’ Zn+2 (aqu.) + Cu. At cathode: Cu+2 + 2e- โ†’ Cu. At anode: Zn ๏‚ฎ Zn2+ + 2 e- . 2) So , since reduction occurs at cathode therefore Cu is โ€“ve and oxidation occurs at anode therefore Zn is +ve. (b) Cu + 2AgNO3 โ†’ Cu (NO3)2 + 2Ag. 1) Cu+2 +Ag โ†’ Ag+ + Cu. At cathode: Cu+2 + 2e- โ†’ Cu At anode: Ag โ†’ Ag+1 + e- 2) So , since reduction occurs at cathode therefore Cu is โ€“ve and oxidation occurs at anode therefore Ag is +ve. (c) Zn + H2SO4 ZnSO4 + H2 1) Zn+2 +H2 2Zn + 2H+ . 2) At cathode: Zn2+ + 2 e- 2 Zn. At anode: H2 2H+ + 2e- . 3) So , since reduction occurs at cathode therefore Zn is โ€“ve and oxidation occurs at anode therefore H is +ve.
  32. 32. Page 32 of 50 Question # 09 What is corrosion? Explain reactions associated with corrosion. (With equations and electrode potential values). OR Define corrosion and explain the reaction involved in rusting of iron with equations and redox potential values. CORROSION Corrosion is an electrochemical reaction composed of two half cell reactions, an anodic reaction and cathodic reaction. In anodic reactions electron releases and in cathodic reactions electron consumes. An electrochemical reaction is defined as a chemical reaction involving the transfer of electrons. It is also a chemical reaction which involves oxidation and reduction. Since metallic corrosion is almost always an electrochemical process, it is important to understand the basic nature of electrochemical reactions. The discoveries that gradually evolved in modern corrosion science have, in fact, played an important role in the development of a multitude of technologies we are enjoying today. An important achievement early in the history of electrochemistry was the production of power sources, following the production of the first batteries by Alessandro Volta illustrates the principle of a Daniel cell in which copper and zinc metals are immersed in solutions of their respective sulfates. The Daniel cell was the first truly practical and reliable electric battery that supported many nineteenth-century electrical innovations such as the telegraph. In the process of the reaction, electrons can be transferred from the corroding zinc to the copper through an electrically conducting path as a useful electric current. Zinc more readily loses electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow through an external wire which leads from the zinc to the copper. The difference in the susceptibility of two metals to corrode can often cause a situation that is called galvanic corrosion named after Luigi Galvani, the discoverer of the effect. The purpose of the separator shown in is to keep each metal in contact with its own soluble sulfates, a technical point that is critical in order to keep the voltage of a Daniel cell relatively constant . The same goal can be achieved by using a salt bridge between two different beakers. The salt bridge in that case, provides the electrolytic path that is necessary to complete an electrochemical cell circuit. This situation is common in natural corrosion cells where the environment serves as the electrolyte that completes the corrosion cell. The conductivity of an aqueous environment such as soils, concrete, or natural waters has often been related to its corrosivity. Zinc anode: Zn(s) โ†’ Zn2๏€ซ๏€ ๏€ซ๏€ 2e- Copper cathode: Cu2๏€ซ๏€ ๏€ซ๏€ 2e- โ†’ Cu(s)
  33. 33. Page 33 of 50 The short-hand description is valid for both cells. Such a description is often used to simplify textual reference to such cells. Zn/Zn2๏€ซ, SO24 (Conc1)//Cu2๏€ซ, SO24 (Conc2))/Cu(๏€ซ) Conc1 andConc2 indicate respectively the concentration of zinc sulfate and copper sulfate that may differ in the two half-cells while the two slanted bars (//) describe the presence of a separator. The same short-hand description also identifies the zinc electrode as the anode that is negative in the case of a spontaneous reaction and the copper cathode as positive. The fact that corrosion consists of at least one oxidation and one reduction reaction is not always as obvious as it is in chemical power cells and batteries. The two reactions are often combined on a single piece of metal as it is illustrated schematically A piece of zinc immersed in hydrochloric acid solution is undergoing corrosion. At some point on the surface, zinc is transformed to zinc ions. This reaction produces electrons and these pass through the solid conducting metal to other sites on the metal surface where hydrogen ions are reduced to hydrogen gas Anodic reaction: Zn(s)โ†’2e- + Zn+2 Cathodic reaction: 2H+2+2e- โ†’ H2(g) The nature of an electrochemical reaction typically illustrated for zinc .During such a reaction, electrons are transferred, or, viewing it another way, an oxidation process occurs together with a reduction process. The overall corrosion processes are summarized in Overall corrosion reaction: Zn ๏€ซ2H๏€ซ ๏€  Zn2๏€ซ๏€ ๏€ซ๏€ H2(g) Briefly then, for corrosion to occur there must be a formation of ions and release of electrons at an anodic surface where oxidation or deterioration of the metal occurs. There must be a simultaneous reaction at the cathodic surface to consume the electrons generated at the anode. These electrons can serve to neutralize must go on at the same time and at equivalent rates. However, what is usually recognized as the corrosion process occurs only at the areas that serve as anodes. Anodic Processes: Let us consider in greater detail what takes place at the anode when corrosion occurs. For instance, reconsider. This re- action involves the reduction of hydrogen ions to hydrogen gas,. This hydrogen evolution reaction occurs with a wide variety of metals and acids, including hydrochloric, sulfuric,
  34. 34. Page 34 of 50 perchloric, hydrofluoric, formic, and other strong acids. The individual anodic reactions for iron, nickel, and aluminum are listed as follows: Iron anodic reaction: Fe(s) Fe2๏€ซ ๏€  ๏€ซ๏€ 2e Nickel anodic reaction: Ni(s) Ni2๏€ซ ๏€  ๏€ซ๏€ 2e Aluminum anodic reaction: Al(s) Al3๏€ซ ๏€  ๏€ซ๏€ 3e That is, the corrosion of metal M results in the oxidation of metal M to an ion with a valence charge of nand the release of n electrons. Some metals such as silver are univalent, while other metals such as iron, titanium, and uranium are multivalent and possess positive charges as high as is general and applies to all corrosion reactions. Cathodic Process When hydrogen ions are reduced to their atomic form they often combine, as shown earlier, to produce hydrogen gas through reaction with electrons at a cathodic surface. This reduction of hydrogen ions at (OH ) ions and make the solution less acidic or more alkaline or basic at the corroding interface. In neutral waters the anodic corrosion of some metals like aluminum, zinc, or magnesium develops enough energy to split water directly. Water splitting cathodic reaction: 2H2O(l) ๏€ซ๏€ 2e- H2 ๏€ซ๏€ 2OH- The change in the concentration of hydrogen ions or increase in hydroxyl ions can be shown by the use of pH indicators, which change color and thus can serve to demonstrate and locate the existence of surfaces on which the cathodic reactions in corrosion are taking place. There are several other cathodic reactions encountered during the corrosion of metals. These are listed below: Oxygen reduction: (acid solutions) O2 ๏€ซ๏€ 4H๏€ซ๏€ ๏€ซ๏€ 4e 2H2O (neutral or basic solutions) O2 ๏€ซ๏€ 2H2O ๏€ซ๏€ 4e 4OH Hydrogen evolution: 2H๏€ซ๏€ ๏€ซ๏€ 2e H2(g) Metal ion reduction: Fe3๏€ซ๏€ ๏€ซ๏€ e Fe2๏€ซ Metal deposition: Cu2๏€ซ๏€ ๏€ซ๏€ 2e Cu(s) Hydrogen ion reduction, or hydrogen evolution, has already been discussed. This is the cathodic reaction that occurs during corrosion in acids. Inhibition of corrosion Introduction: Inhibition means to remove corrosion. There are a number of ways to remove corrosion. Some methods are: ๏‚ท Impermeable layer. ๏‚ท Galvanizing ๏‚ท Cathodic production. ๏‚ท Sacrification method.
  35. 35. Page 35 of 50 Impermeable layer: Coating of impermeable layer on the material prevents it from corrosion. Sometimes this measure fails because of porous paint. Galvanizing: Coating of zinc on a material can also prevent it from being corroded. This method is called Galvanizing Cathodic production: This method was Invented in 1824 by the UK scientist Sir Humphrey Davy (1778-1829) to protect the copper-cladded wooden ships from seawater corrosion. It is an electrical method of preventing corrosion on metallic structures which are in electrolytes such as soil or water Sacrification method: This method is basically used in ships. Here Mg is used as a sacrificing agent. We attach Mg to Fe.First Mg oxidizes, and then Fe will oxidize after a long time. Question # 10 Write three definitions of oxidation and reduction reactions with examples ; also explain second law of electrolysis. Definitions of oxidation and reduction reactions: Reduction reaction: (1) Loss of O2 (2) Increase of H+ ion. (3) Increase of e- . (4) Increase of oxidation no. Examples: Zn ๏‚ฎ Zn2+ + 2 e- . Zn is the reducing agent, and Zn2+ the oxidizing agent. Oxidation reaction: (1) Gain of O2. (2) Decrease of H+ ion. (3) Decrease of e- . (4) Decrease of oxidation no. Examples: Cu2+ + 2 e ๏‚ฎ Cu Cu2+ is the oxidizing agent and Cu the reducing agent.
  36. 36. Page 36 of 50 Faradayโ€™s Second Law of Electrolysis: โ€œIt states that masses of different substances deposited or liberated when same quantity of current is passed through different electrolytes connected in series are proportional to their chemical equivalent massesโ€ Explanation: Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of electricity through them for the same time. Now Ag Cu and Al metals collect at the cathode. Their masses are directly proportional to their equivalent masses. According to Faraday,if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes, we get which are the equivalent masses of Ag, Cu and Al respectively. Ag+ 1e- โ†’ Ag(S) Cu2+ + 2e- Cu (S) Al+3 +3e- Al(S) Charge on 1 electron=96500 coulomb= 1 faraday
  37. 37. Page 37 of 50 Question # 11 State the methods of prevention of corrosion by sacrificial anode method with diagram. Corrosion can be prevented by application of electrochemistry principles. This basically falls into two distinct areas,sacrificial anodes and cathodic protection by impressed currents. Sacrificial Anodes. In this preventative technique, corrosion is allowed to occur on a piece of metal that is extraneous to the structure, for example, a zinc attached to a steelboat hull. The zinc corrodes in place of the steel hull. The principles behind this process were discussed previously and will not be repeated except to show the relevant Evans diagram. Cathodic Reaction 1 Anodic Reaction 2 icorr 1+2 Ecorr 1+2 E (V) log Current Density ๏ญA/cm 2 Cathode Reaction 3 Anode Reaction 3 Total Cathode 1+3 In this case the anode#3 was protected from corrosion by anode reaction #2. One initial principle is that the sacrificial material must have a potential lower than the material it is trying to protect. Simple examples of the application of this protection technique include:- Galvanized bolts, automobile steel and mail boxes, zincs placed outboard engines and steelboat hulls, aluminum blocks on oil rigs, etc. Typical coatings which work on this principle are:- Zinc on steel, aluminum on steel, cadmium on steel. It should be remembered that cadmium is only slightly below steel in the galvanic series. As such it does not have much "Throwing power" which is the ability to protect over large distances.
  38. 38. Page 38 of 50 Question # 12 Consider the reaction 2Ag+ + Cd 2Ag + Cd+2 Standard electrode potentials are Eo Ag + /Ag =0.80 volts and E0 Cd +2 /Cd = - 0.40 volts. (1) What is the standard electrode potential (Eo ) for this reaction. Eo = E - E = EREDUCTION - EOXIDATION = 0.80 โ€“ (-0.4) EO = 1.20 volts. (2) Which electrode is negative electrode? ๏‚ท The electrode which has low redox potential value is โ€“ve electrode. ๏‚ท In this case Cd has low value of redox potential i.e -0.40 volts. So , Cd is โ€“ve. ๏‚ท Question # 13: What is redox potential? How will you obtain redox potential of a complete cell? Redox potentials are used to infer the direction and free energy cost of reactions involving electron transfer,one of the most important types of biochemical reactions. Such reduction-oxidation reactions are characterized by a free energy change. If we are given two half cell reactions with there respective redox potential values then e.g ๏‚ท CO+2 (aqu.) + 2e- CO -0.28 volts. ๏‚ท Br2(aqu.) + 2e- 2Br- (aqu.) 1.07 volts. - - - - 1.35 volts. Then, redox potential values add up to 1.35 volts.this is how we obtain redox potential value of a complete cell. Question # 14 Consider equations with reduction potentials and answer the followings. E/V O2 (g) +4H+ (aqu.) + 4e- 2H2O (l) +1.23 Cl2 (g) + 2e- 2Cl- (aq) +1.36 F2 (g) +2e- 2F- (aqu.) +2.87
  39. 39. Page 39 of 50 1-State and explain what occurs when Flourine is bubbled into water. When fluoride is bubbled in into the water then after these standard reactions, By applying anticlockwise rule. O2 (g) +4H+ (aqu.) + 4e- 2H2O (l) +1.23 2F2 (g) +4e- 4F- (aqu.) +2.87 - - - - O2 + 4H+ (aqu.) + 4F(aqu.) - 2H2O + 2F2 -1.64 volts. So, ๏‚ท O2 is formed when F2 is bubbled in water H2O. ๏‚ท Water is oxidized and fluorine is reduced. (1) Write an equation for the expected reaction of chlorine with water using the above data. O2 (g) +4H+ (aqu.) + 4e- 2H2O (l) +1.23 2Cl2 (g) + 4e- 4Cl- (aq) +1.36 - - - - O2 + 4H+ (aqu.) + 4Cl- (aqu.) 2H2O + 2F2 -0.13 volts. So, ๏‚ท O2 is formed. ๏‚ท Water is oxidized and chlorine is reduced. (2) In fact the reaction between chlorine and water is usually represented by the equation. Cl2(g) + H2O(l) H+ (aqu.) + Cl- (aqu.) + HClO(aqu.) Suggest why this reaction occurs rather than that in (2). ๏‚ท This is because of little difference between redox potential values. ๏‚ท Cl2 mix with water to form bleaching agent. Question # 15 A direct current of 0.01 amp flows for 4 hours through three cells in series. They contain aqueous AgNO3 , CuSO4 and Au(NO3)3. Calculate the mass of metal deposited in each. (atomic masses. Ag=108, Cu=63.5, Au=197 respectively). Step#1: Find Charge OR Coulomb q=It = (0.01)(4 x 60 x 60) q= 144 coulomb Step #2: Find Faraday 96500 coulomb = 1 Faraday 144 coulomb = 1/96500 x 144 F= 1.49 x 10-3 Farad.
  40. 40. Page 40 of 50 Step #3: Taking the calculated values of Charge and faraday,solve the Question further for given equations Let X be the required amount of AgNO3,CuSO4 and AuNO3. Ag+ 1e- โ†’ Ag(S) 108 g = 1 F 1.49 x 10-3 = X gm X= 0.161 gm. Cu2+ + 2e- โ†’ Cu(S) 63.5/2= 1 F OR 2 F= 63.5 gm 1.49 x 10-3 = X 2X = 63.5 x (1.49 x 10-3)/2 X= 0.0473 gm. Al+3 +3e- ๐ด๐‘™(๐‘ ) 197 gm = 3 F 3 F= 197 gm 1.49 x 10-3 F= X 3X= 197 x (1.49 x 10-3)/3 X= 0.0983 gm Question # 16 A metal of Ar = 27 is deposited by electrolysis when 0.15amp flows for 3.5 hrs,0.176gm of metal is deposited. What is the charge on the cation of these metals? Step#1: Find Charge OR Coulomb q=It =0.15 (3.5x60x60) q= 1890 coulomb Step #2: Find Faraday 96500 coulomb = 1 faraday 1890 coulomb = x x= 1890/96500 x= 0.0196F Step#3: 0.176gm= 0.0196F 27gm = x faraday x = 0.0196x27/0.176 x = 3 charge is +3.
  41. 41. Page 41 of 50 1. IMPORTANT KEY POINTS AND QUESTIONS. Questions: (1) Pak has crisis of electricity. U.P.S is used. State two laws of cell used in U.P.S.? Answer: (1) Faraday's first law of electrolysis: Statement: โ€œThe amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of current passing through the electrolyteโ€ Explanation: It means that the mass of a substance produced by electrolysis is proportional to the quantity of electricity used. In order to produce more substance electrolytically we must use more electricity. m ฮฑ Q m ฮฑ It m=zIt where โ€œZโ€ is known as electrochemical equivalent. Definition of Electrochemical Equivalent โ€œzโ€: โ€œIf the mass is 1 kg and current of 1 ampere is passed through electrolyte in 1 second then the mass deposited or liberated is called 1 electrochemical equivalent โ€ OR โ€œThe mass deposited by 1 coulomb charge is called 1 electrochemical equivalentโ€ z=m / It or z=m/Q For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited is 0.001118 g. this is the value of electrochemical equivalent of silver. Importance offaradayโ€™s First law ofElectrolysis: With the help of faradayโ€™s First law we are able to calculate: 1) The value of electrochemical equivalents of different substances. 2) The masses of substances produced by passing a known quantity of electricity through their solution. (2) Faradayโ€™s Second Lawof Electrolysis: โ€œIt states that masses of different substances deposited or liberated when same quantity of current is passed through different electrolytes connected in series are proportional to their chemical equivalent massesโ€
  42. 42. Page 42 of 50 Explanation: Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of electricity through them for the same time. Now Ag Cu and Al metals collect at the cathode. Their masses are directly proportional to their equivalent masses. According to Faraday,if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes, we get which are the equivalent masses of Ag, Cu and Al respectively. Ag+ 1e- Ag(S) Cu2+ + 2e- Cu (S) Al+3 +3e- Al(S) Charge on 1 electron = 96500 coulomb = 1 faraday Equivalent Mass: โ€œThe mass deposited or liberated which is equal to 108 parts of silver, 35.5 parts of chlorine and 8 parts of oxygenโ€ . Importance ofFaradayโ€™s Second Law: By the help of 2nd law we can calculate: ๏‚ท Equivalent weight of metal. ๏‚ท Unit of electric charge (coulomb). ๏‚ท Avogadroโ€™s number. (2) Rate of corrosion depends on surface exposure? Answer:please paste โ€˜inhibition of corrosionโ€™ here. (3) Define cathodic reaction , corrosion and equation with redox potential values? Answer:please paste from โ€˜corrosion to inhibitionโ€™ here. Important points: Carbon atom has no charge. Charge on carbon will depned upon with what species it is attached to, if it is attached to more electronegative element than itself, then the charge should be +ve and if it is attached to more electropositive element then the charge should be -ve on carbon. Although in general carbon forms covalent bonds most of the time so it does not form ionic species like 4+ (by loss of 4 electrons) or 4- (by gain of 4 electrons). It shares its 4 electron with 4 others to form covalent bonds.
  43. 43. Page 43 of 50 (1) If metal is reduced then redox potential value is more. (2) If metal is oxidized then redox potential value is small. When there is reduction there will be deposition always on cathode. When battery is attached then anode = +ve and cathode = -ve and if battery is not attached then anode = -ve and cathode = +ve. The electrode which has low redox potential value is โ€“ve electrode. In cathodic reaction electron consumes whereas in anodic reaction electron releases. Whenever you have ions always write the state. E.g :- Cl2(g) + H2O(l) H+(aqu.) + Cl-(aqu.) + HClO(aqu.) Gold has high redox potential value so it does not corrode.
  44. 44. Page 44 of 50 5. M.C.Qโ€™s RELATING TO ELECROCHEMISTRY (1). Faradayโ€™s laws of electrolysis are related to the a. atomic number of the cation. b. atomic number of the anion. c. equivalent weight of the electrolyte d. speed of the cation Ans: (c) (2). A solution containing one mole for litre each of Cu(NO3)2, AgNO3, Hg2(NO3)2 and Mg(NO3)2 is electrolysed using inert electrodes. Standard electrode potentials in volts (reduction potentials) are: Ag+| Ag = 0.80 Hg22+ |Hg = 0.79 Cu2+|Cu = 0.34 Mg2+|Mg = -2.37 With increasing voltage, the sequence of deposition of metals on the cathode will be: a. Ag,Hg,Cu, Mg b. Mg,Cu,Hg,Ag c. Ag,Hg, Cu d. Cu,Hg, Ag Ans: ( c ) (3). The electric charge or electrode deposition of one gram equivalent of a substance is: a. one ampere for one second b. 96,500 coulombs per second c. one ampere for one hour d. charge on one mole of electrons Ans: (d) (4). The reaction ยฝ H2(g) + AgCl(s) = H+(aq) + Cl- (aq) +Ag(s) occurs in the galvanic cell: a. Ag|AgCl(s) |KCl(soln) |AgNO3(soln) |Ag b.Pt|H2(g) |HCl(soln) |AgNO3(soln) |Ag c. Pt|H2(g) |HCl(soln) |AgCl(s) |Ag d. Pt|H2(g) |KCl(soln) |AgCl(s) |Ag
  45. 45. Page 45 of 50 Ans: (c) In cell (c) reactions occurring are AgCl(s) + e- -> Ag(s) + Cl- (aq) ยฝ H2(g) -> H+ (aq) + e- adding ยฝ H2(g) + AgCl(s) = H+(aq) + Cl-(aq) +Ag(s) (5) Choose the single correct or the single incorrect statement A. At the anode, oxidation takes place only when used as a battery B. At the anode, oxidation takes place in a battery and in a electrolysis operation C. At the cathode, oxidation takes place only when used as a battery D. At the cathode, oxidation takes place in a battery and in a electrolysis operation Answer B. Consider... It is a convention to call the reduction electrode a cathode in a battery or in a electrolysis operation. The oxidation reaction occurs at the anode. (6)All chemical reactions that supply the power to a battery are oxidation reduction reactions. True or false? Answer True! Consider... Only RedOx reactions involve electron transfer. Even concentration cells involve oxidation and reduction of the same material. Note that Ag+ + Cl- AgCl(s) is an ionic reaction, not a redox reaction. (7)The half-cell using the reaction: 2 H+ (aq, 1.00 F) + 2 e ยฎ H2(g, 1atm) has a half cell potential of zero because E. it is so defined, F. hydrogen is not very reactive, G. its potential is absolutely zero, H. it is not a useful electrode. Answer A Consider... No cell potential is ABSOLUTELY zero. H2 is reactive. This is not the reason at all. (8)The notation to indicate a boundary between two phases in a electro- chemical cell is I. | J. / K. || L. Answer A. Consider... The vertical bar | is used to indicate boundary between two phases. Pt | H2 | H+ (1.0 M) represents the hydrogen half cell. (9)The notation to indicate a salt bridge between two half electro- chemical cells is M. |
  46. 46. Page 46 of 50 N. / O. || P. Answer C. Consider... Only | and || are used among the four notations. Two vertical bars, ||, represent a salt bridge. 10. Electricity can pass through molten lead(II) bromide because of the presence of a. free electrons b. moveable ions c. moveable atoms d. lead metal 11. When a dilute salt water is electrolysed, a colorless gas is given off at the anode. The gas is a. hydrogen b. steam c. oxygen d. chlorine 12. A solution of copper(II) sulphate is electrolysed, using carbon electrodes. The pinkish deposit which forms on one of the electrodes is a. copper b. copper(I) oxide c. copper(II) oxide d. copper(III) sulphide 13. A solution of copper(II) sulphate is electrolysed, using copper electrodes. Which of the following would happen? a. the anode loses weight b. the cathode loses weight c. the solution darkens in color d. the solution lightens in color 14. An electrolyte is always a. an acid or alkali b. an aqueous solution c. a liquid d. a molten solid 15. Anions are formed by a. metals gaining electrons b. metals losing electrons c. non-metals gaining electrons d. non-metals losing electrons 16. Which of these anions is never discharged at the positive electrode during electrolysis? a. NO3- b. OH- c. I- d. O2-
  47. 47. Page 47 of 50 17. In the electrolytic manufacture of aluminium, what is the anode made of? a. copper b. graphite c. platinum d. steel 18. In which electrolyte would a carbon cathode increase in mass during electrolysis? a. aqueous copper(II) sulphate b. concentrated hydrochloric acid c. concentrated aqueous sodium chloride d. dilute sulphuric acid 19. Chlorine is manufactured commercially by the electrolysis of aqueous sodium chloride (brine). Which other important products are made in the process? a. hydrochloric acid and hydrogen b. hydrogen and sodium c. hydrogen and sodium hydroxide d. sodium and sodium hydroxide 20. An electric current is passed through aqueous potassium sulphate, K2SO4. What is formed at the cathode (negative electrode)? a. hydrogen b. oxygen c. potassium d. sulphur 21. What happens when molten lead(II) chloride is electrolysed? a. chloride ions gain electrons at the cathode b chloride ions lose electrons at the anode c. lead(II) ions lose electrons at the cathode d. lead(II) ions move towards the anode 22. Which element is liberated at a carbon cathode when aqueous sodium chloride is electrolysed? a. chlorine b. hydrogen c. oxygen d. sodium 23. Which change always takes place when aqueous copper(II) sulphate is electrolysed?
  48. 48. Page 48 of 50 a. copper is deposited at the negative electrode b. oxygen is evolved at the positive electrode c. sulphate ions move towards the negative electrode d. the color of the solution fades 24. Which element is liberated at the cathode by the electrolysis of an aqueous solution containing its ions? a. bromine b. chlorine c. hydrogen d. oxygen 25. Aqueous copper(II) sulphate is electrolysed using copper electrodes. Which observations will be made? at anode (positive) at cathode (negative) a anode dissolves pink solid forms b anode dissolves pink solid forms c color gas forms color gas forms d color gas forms pink solid forms 26. Why is cryolite, Na3AlF6, used in the extraction of aluminium from aluminium oxide? a. to dissolve aluminium oxide b. to prevent the anodes from burning away c. to prevent the oxidation of the aluminium d. to remove impurities from the aluminium oxide 27. When sodium chloride was electrolysed, sodium was produced at the negative electrode. In which form was the sodium chloride during the electrolysis? a. concentrated aqueous solution b. dilute aqueous solution c. molten d. solid 28. In which instance is there no change in the concentration of the solution during electrolysis? a. concentrated sodium chloride solution between carbon electrodes b. copper(II) sulfate solution between copper electrodes c. copper(II) sulfate solution between platinum electrodes d. dilute sodium chloride solution between platinum electrodes
  49. 49. Page 49 of 50 29. An example of a weak electrolyte is a. alcohol b. salt solution c. sugar solution d. ammonia solution 30. Electroplating iron with zinc is called galvanising. The reaction at the cathode is shown by the equation a. Fe (s) ---> Fe2+ (aq) + 2e- b. Fe2+ (aq) + 2e- ---> Fe (s) c. Zn (s) ---> Zn2+ (aq) + 2e- d. Zn2+ (aq) + 2e- ---> Zn (s) 31. The circuit shown below was set up, with brass as the anode. Which electrode reactions will occur on closing the switch? Anode reaction Cathode reaction a. Copper dissolves preferentially. Copper is deposited. b. Copper dissolves preferentially. Hydrogen is evolved. c. Zinc dissolves preferentially. Hydrogen is evolved. d. Zinc and copper both dissolve. Copper is deposited. 32. During the electrolysis of concentrated sodium chloride in a cell, chlorine, hydrogen, and sodium hydroxide are produced. What is the molar ratio of these products? Chlorine Hydrogen Sodium hydroxide a. 1 1 1 b. 2 1 2
  50. 50. Page 50 of 50 c. 2 1 1 d. 2 2 1 Answers 10. b 11. c 12. a 13. a 14. c 15. c 16. a 17. b 18. a 19. c 20. a (H+ and K+ ions in the electrolyte migrate to the cathode. H+ are preferentially discharged to form hydrogen gas because it is lower down in the electrochemical series than K+ ions) 21. b (the negative chloride ions will migrate to the anode and become oxidised at the anode to form chlorine gas) 22. b (the ions attracted to the cathode are H+ and Na+ ions. H+ is preferentially discharged to form hydrogen gas) 23. b 24. c 25. b 26. a 27. c 28. b 29. d 30. d 31. c 32. b

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